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I know this not that well posed of a question so please bear with me. Suppose we have a $n$-dimensional psuedo-Riemannian manifold $(M,g)$. We have that there are $n^2$ functions that make up $g=[g_{ij}]$. However not all of them can be chosen arbitrarily, since $g_{ij}=g_{ji}$. So that leaves $\frac{n(n+1)}{2}$ functions remaining. Now my question for an arbitrary $g$, how many of those functions are truly "independent", and determine $g$.

The reason I ask is because I was looking at the case where $n=2$. One can always find isothermal coordinates for $(M,g)$ in which $g_{ij}=\Lambda\delta_{ij}$, where $\Lambda$ is a smooth function and $\delta_{ij}$ is the kronecker delta. So in the case of $n=2$ we have that $(M,g)$ has 1 degree of freedom.

I have heard in passing that in general that there are $\frac{n(n-1)}{2}$ degrees of freedom. If someone could help me either word this better, give me resources or offer a heuristic argument as to why that is true it would be great.

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    $\begingroup$ bare means to take off your clothes. bear means to be patient, so 'bear with me' is what you meant. Probably related to the word burden. $\endgroup$ – Will Jagy Dec 18 '13 at 21:04
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    $\begingroup$ noted and fixed :) $\endgroup$ – Saif Dec 18 '13 at 21:05
  • $\begingroup$ You might want to look at Riemann's inaugural lecture, in which he addressed almost exactly this question (although for the Riemannian case rather than pseudo-Riemannian). It's transcribed in vol 3 of Spivak's differential geometry book, and then the next few chapters try to explain what the heck Riemann was saying. It's pretty good reading, no matter what. $\endgroup$ – John Hughes Dec 18 '13 at 21:24
  • $\begingroup$ Thank you I will check it out later tonight when I go to the library. Do you recall the gist of his argument? $\endgroup$ – Saif Dec 18 '13 at 21:29
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The answer is quite trivial. In $n$ dimensions you are free to change the coordinates by providing $n$ independent functions. So the actual number of the degrees of freedom for the metric is $$n(n+1) / 2 - n = n(n-1) /2.$$ The answer, which only depends on the general rank discussion, is therefore the same for the pseudo-metric as well.

You have implicitly mentioned this fact already for $n=2$ when you said one can find isothermal coordinates. This is essentially nothing more than diagonalization procedure. What can be surprising is that for $n > 2$ the diagonalization can fail for some metrics. Even though from linear algebra we know that the metric can be diagonalized at every point, over bigger portions of the manifold this need not be the case.

In the general theory of relativity this phenomenon manifests as a frame dragging caused by the $g_{\phi t}$ term of the metric.

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  • $\begingroup$ I know this question is old, but I wondered: Do you have a precise statement for the idea that the "metric, up to coordinate changes, really depends only on $\frac{n(n+1)}{2}-n$ functions"?. Of course, I see where the $\frac{n(n+1)}{2}$ comes from (this is the dimension of symmetric matrices). Is there a formal statement, that gives you the number $\frac{n(n+1)}{2}-n$ degrees of freedom? $\endgroup$ – Asaf Shachar May 31 '18 at 9:00
  • $\begingroup$ @AsafShachar This comes from the contracted Bianchi Identities of which there are $n$ in $n$-dimensions. $\endgroup$ – Sam Colley Jun 28 '19 at 10:34

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